Question

If theta is an acute angle and tan theta=1/√7then find the value of cos es^2theta-sec^2theta/cos es^2theta+sec^2theta.

Answer 1

✪ANSWER✪

• $\boxed{ \frac{cosec^2(\theta) - sec^2(\theta)}{cosec^2(\theta) + sec^2(\theta)} = \frac{3}{4}}$

★GIVEN★

• $tan(\theta) = \frac{1}{\sqrt{7}}, 0 ≤\theta ≤ 90^0.$

⍟CALCULATION⍟

Here,

$\to \frac{cosec^2(\theta) - sec^2(\theta)}{cosec^2(\theta) + sec^2(\theta)}$

$\to \frac{\frac{1}{sin^2(\theta)} - \frac{1} {cos^2(\theta)}}{\frac{1}{sin^2(\theta)} + \frac{1} {cos^2(\theta)} }$

$\to \frac{sin^2(\theta)\left[\frac{1}{sin^2(\theta)} - \frac{1} {cos^2(\theta)}\right]}{sin^2(\theta)\left[\frac{1}{sin^2(\theta)} + \frac{1} {cos^2(\theta)}\right]}$

$\to \frac{1 - \frac{sin^2(\theta)}{cos^2(\theta)}}{1 + \frac{sin^2(\theta)}{cos^2(\theta)}}$

$\to \frac{1 - tan^2(\theta)}{1 + tan^2(\theta)}$

$\to \frac{1 - \left(\frac{1}{\sqrt{7}}\right)^2}{1 + \left(\frac{1}{\sqrt{7}}\right)^2}$

$\to \frac{1 - \frac{1}{7}}{1 + \frac{1}{7}}$

$\to \frac{7 - 1}{7 + 1}$

$\to \frac{6}{8}$

$\to \frac{3}{4}$